复数的乘法恒等式和乘法倒数是什么？

复数是 实数和虚数之和。这些数字可以写成 a+ib 的形式，其中 a 和 b 都是实数。它用 z 表示。

在复数形式中，值“a”称为实部，用 Re(z) 表示，“b”称为虚部 Im(z)。在复数形式中，a +bi 'i' 是一个称为“iota”的虚数。

实数

数字表示一个数制，如正数、负数、零、整数、有理数、无理数、分数等，称为实数。这些表示为 Re()。

例如：14、-45、0、1/7、2.8、√5等，都是实数。

虚数

非实数称为虚数。对一个虚数进行平方后，结果为负。虚数表示为 Im()。

示例：√-5、√-7、√-11 都是虚数。这里的“i”是一个名为“iota”的虚数。

复数的乘法恒等式和乘法倒数是什么？

解决方案：

Multiplicative identity of complex number says that when the complex number is multiplied by the number 1 it will give that number as product.

 “1” is the multiplicative identity of a number.  If the complex number being multiplied is 1 itself. 

The multiplicative identity property of complex number is represented as: z.1 = z = 1.z

For example: Let assume z = a+ib, then as per the property z.1 = z

therefore, a+ib.1+0i 

             = a + 0i +bi + 0i 

             = a +bi

             = z . for all z ∈ C.

Multiplicative inverse of any number N is represented by 1/N or N^-1. It is also called reciprocal of number. The reciprocal of a number is that when it is multiplied with the original number the value equals to identity 1, or you can say that it is a method of dividing a number by its own to generate identity 1, such as N/N = 1.

When a number is multiplied by its own multiplicative inverse the resultant value is equal to 1. The multiplicative inverse of a complex number z is simply 1/z. It is denoted as:

1/z or z^-1 (Inverse of z)

It is also called as the reciprocal of a complex number and 1 is called the multiplicative identity..

For example: Let assume z = a+ib, then as per the property z.1 = z, its inverse is z = 1/z

multiplicative inverse of z = a +ib = 1/a+ib 

                                         = 1/(a+ib) x (a-ib)/(a-ib)

                                         = (a-ib ) / (a² -b² i² )

                                         = (a-ib) / (a² + b²)

                                         = a / (a²+b²) – {b /(a²+b²)} i

示例问题

问题 1：求 -3 + 8i 的乘法逆元？

解决方案：

The multiplicative inverse of a complex number z is simply 1/z. 

It is denoted as : 1 / z  or  z^-1 (Inverse of z)

         here z = -3 +8i 

Therefore z = 1/z

                  = 1 / (-3 +8i) 

Now rationalizing 

                   = 1/(-3+8i) x (-3-8i)/(-3-8i)

                   = (-3-8i) / {(-3)² – 8²i²}

                   = (-3-8i) / {9 +64}

                   = (-3-8i)/ (73)

                   = -3/73 – 8i/73

问题 2：求 2 – 3i 的乘法逆元？

解决方案：

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as : 1 / z  or  z^-1 (Inverse of z)

        here z = 2 – 3i

Therefore z = 1/z

                  = 1 / (2 – 3i)

Now rationalizing

                  = 1/(2 – 3i) x (2 + 3i)/(2 +3i)

                  = (2 + 3i) / {(2)2 – 3²i²}

                  = (2 + 3i) / { 4 + 9}

                  = (2 + 3i)/ (13)

                  = 2/13 + 3i/13

问题 3：求 √5+3i 的乘法逆元？

解决方案：

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as: 1 / z  or  z^-1 (Inverse of z)

       here z = √5+3i

Therefore z = 1/z

                  =  1 / (√5+3i )

now rationalizing

                 = 1/(√5+3i ) x (√5 -3i)/(√5-3i)

                 = (√5+3i ) / {(√5)² – (3)² (i)² }

                 =  (√5 +3i) / { (5) + 9}

                 =  (√5 + 3i)/ (14)

                 = √5/14 + 3i/14

问题 4：求 4 – 3i 的乘法逆元？

解决方案：

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as : 1 / z  or  z-1 (Inverse of z)

       here z = 4 – 3i

therefore z = 1/z

                 =  1 / (4 – 3i)

Now rationalizing

                 = 1/(4 – 3i) x (4 + 3i)/(4 +3i)

                 = (4 + 3i ) / {(4)² – 3²i²}

                 = (4 + 3i) / {16 + 9}

                 = (4 + 3i)/ (25)

                 = 4/25 + 3i/25

问题 5：求 (5-7i) 的乘法逆元？

解决方案：

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as : 1 / z  or  z-1 (Inverse of z)

      here z = 5 – 7i

therefore z = 1/z

                 = 1 / (5 – 7i)

Now rationalizing

= 1/(5 – 7i) x (5 + 7i)/(5 +7i)

= (5 + 7i ) / {(5)² – 7²i²}

= (5 + 7i) / { 25 + 49}

= (5 + 7i)/ (74)

= 5/74 + 7i/74

问题 6：化简 (2-4i)(5-7i) 并找到它的乘法逆元？

解决方案：

Given:  (2-4i)(5-7i)

= 10 -14i -20i +28i2

= 10 -14i -20i + 28(-1)2

= 10 – 14i – 20i +28

= 18 – 34i

Now, multiplicative inverse of 18 – 34i is

It is denoted as : 1 / z  or  z-1 (Inverse of z)

Here z = 18 – 34i

Therefore z = 1/z

                 =  1 / (18 – 34i)

Now rationalizing

= 1/(18 – 34i) x (18 + 34i)/(18 + 34i)

= (18 + 34i) / {(18)² – 34²i²}

= (18 + 34i) / {324 + 1156}

= (18 + 34i)/ (1480)

= 18/1480 + 34i/1480

= 9/740 + 17i/740

问题 7：求 (4 + 2i) 的乘法逆元？

解决方案：

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as : 1 / z  or  z-1 (Inverse of z)

       here z = 4 + 2i

Therefore z = 1/z

                  =  1 / (4 + 2i)

Now rationalizing

= 1/(4 + 2i) x (4 – 2i)/(4 – 2i)

= (4 – 2i) / {(4)² – 2²i²}

= (4 – 2i) / {16 + 4}

= (4 – 2i)/ (25)

= 4/25 – 2i/25